Trigonometry puzzle

  • Thread starter arildno
  • Start date
  • #1
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131

Main Question or Discussion Point

Trigonometry "puzzle"

If we know the perimeter P of a triangle, a side S of it, and the height H down on the line going through S, then we have uniquely determined the triangle.

It is not wholly trivial to derive the expressions for the other two sides in terms of P,S and H.
 

Answers and Replies

  • #2
664
3
I'm forgetting my algebra, but:

1) Essentially, we are given an ellipse, with the focii seperated by S, a major axis of P, and if expressed as y=f(x), the corresponding y value of H for which we must solve.

2) The ellipse as a function (not purely as a function of x) might be:
sqrt((x+S/2)^2 + y^2) + sqrt((x-S/2)^2 + y^2) = P - S

3) Solve for x, given that y=H:
sqrt((x+S/2)^2 + H^2) + sqrt((x-S/2)^2 + H^2) = P - S
(I forget exactly how to go about this...)

4) With x in hand, it becomes trivial. The other sides would be:
Side 1 = sqrt((x+S/2)^2+y^2)
Side 2 = sqrt((x-S/2)^2+y^2)

DaveE
 
Last edited:
  • #3
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
The neat touch is, of course, to go over to a coordinate representation of the end points and utilize the properties of an ellipse.

The "brute force" approach to directly relate the SIDE LENGTHS to each other, say with a couple of Pythagorases will lead to ugly non-linear equations.

EDIT
NOTE:
Your initial formula is wrong. On your right-hand side, it should say P-S, not P!
(The sum of the two other sides is constant)
 
Last edited:
  • #4
664
3
The neat touch is, of course, to go over to a coordinate representation of the end points and utilize the properties of an ellipse.
Heh, that's all well and good if you remember your ellipse properties, but for those of us who're 14 years out of their geometry classes, we're stuck with brute force :) (short of going and looking up all those old properties)

The "brute force" approach to directly relate the SIDE LENGTHS to each other, say with a couple of Pythagorases will lead to ugly non-linear equations.
Out of curiosity, how *would* you go about solving that ugly equation for x? Anyone?

Your initial formula is wrong. On your right-hand side, it should say P-S, not P!
Ooops, fixed!

DaveE
 
  • #5
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
Hmm..you gave the simple ellipse approach in your initial post.
Now, as for solving for x, the simplest way is like this:
[tex]\sqrt{(x-\frac{S}{2})^{2}+H^{2}}=(P-S)-\sqrt{(x+\frac{S}{2})^{2}+H^{2}}[/tex]
Square this expression, gaining:
[tex](x-\frac{S}{2})^{2}+H^{2}=(x+\frac{S}{2})^{2}+(P-S)^{2}-2(P-S)\sqrt{(x+\frac{S}{2})^{2}+H^{2}}[/tex]
Simplify to:
[tex]\sqrt{(x+\frac{S}{2})^{2}+H^{2}}=\frac{xS}{(P-S)}+\frac{(P-S)}{2}[/tex]
Re-square:
[tex](x+\frac{S}{2})^{2}+H^{2}=\frac{S^{2}}{(P-S)^{2}}x^{2}+xs+\frac{(P-S)^{2}}{4}[/tex]
Simplify this to:
[tex](1-\frac{S^{2}}{(P-S)^{2}})x^{2}=\frac{P(P-2S)}{4}-H^{2}[/tex]
From which we gain the positive solution:
[tex]x=\frac{(P-S)}{2}\sqrt{1-\epsilon},\epsilon=\frac{4H^{2}}{P(P-2S)}[/tex]
 
  • #6
Cool question man.
 

Related Threads for: Trigonometry puzzle

  • Last Post
2
Replies
31
Views
3K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
14
Views
6K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
Top